Is sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ) ?

The answer to whether sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ) is True

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prove sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ)

prove

sin (3 θ) = 3 cos^{2} (θ) sin (θ) - sin^{3} (θ)

True

Solution steps

sin (3 θ) = 3 cos^{2} (θ) sin (θ) - sin^{3} (θ)

Manipulating left side

sin (3 θ)

Rewrite using trig identities

sin (3 θ)

Rewrite as

= sin (2 θ + θ)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (2 θ) cos (θ) + cos (2 θ) sin (θ)

Use the Double Angle identity:

sin (2 θ) = 2 sin (θ) cos (θ)

= cos (2 θ) sin (θ) + cos (θ) 2 sin (θ) cos (θ)

Use the Double Angle identity:

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

= (cos^{2} (θ) - sin^{2} (θ)) sin (θ) + 2 cos (θ) cos (θ) sin (θ)

= (cos^{2} (θ) - sin^{2} (θ)) sin (θ) + 2 cos (θ) cos (θ) sin (θ)

Expand

(cos^{2} (θ) - sin^{2} (θ)) sin (θ) + 2 cos (θ) cos (θ) sin (θ) : - sin^{3} (θ) + 3 cos^{2} (θ) sin (θ)

(cos^{2} (θ) - sin^{2} (θ)) sin (θ) + 2 cos (θ) cos (θ) sin (θ)

2 cos (θ) cos (θ) sin (θ) = 2 cos^{2} (θ) sin (θ)

2 cos (θ) cos (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 2 cos^{1 + 1} (θ) sin (θ)

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (θ) sin (θ)

= sin (θ) (cos^{2} (θ) - sin^{2} (θ)) + 2 cos^{2} (θ) sin (θ)

= sin (θ) (cos^{2} (θ) - sin^{2} (θ)) + 2 cos^{2} (θ) sin (θ)

Expand

sin (θ) (cos^{2} (θ) - sin^{2} (θ)) : cos^{2} (θ) sin (θ) - sin^{3} (θ)

sin (θ) (cos^{2} (θ) - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = sin (θ), b = cos^{2} (θ), c = sin^{2} (θ)

= sin (θ) cos^{2} (θ) - sin (θ) sin^{2} (θ)

= cos^{2} (θ) sin (θ) - sin^{2} (θ) sin (θ)

sin^{2} (θ) sin (θ) = sin^{3} (θ)

sin^{2} (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= sin^{3} (θ)

= cos^{2} (θ) sin (θ) - sin^{3} (θ)

= cos^{2} (θ) sin (θ) - sin^{3} (θ) + 2 cos^{2} (θ) sin (θ)

Simplify

cos^{2} (θ) sin (θ) - sin^{3} (θ) + 2 cos^{2} (θ) sin (θ) : - sin^{3} (θ) + 3 cos^{2} (θ) sin (θ)

cos^{2} (θ) sin (θ) - sin^{3} (θ) + 2 cos^{2} (θ) sin (θ)

Group like terms

= - sin^{3} (θ) + cos^{2} (θ) sin (θ) + 2 cos^{2} (θ) sin (θ)

Add similar elements:

cos^{2} (θ) sin (θ) + 2 cos^{2} (θ) sin (θ) = 3 cos^{2} (θ) sin (θ)

= - sin^{3} (θ) + 3 cos^{2} (θ) sin (θ)

= - sin^{3} (θ) + 3 cos^{2} (θ) sin (θ)

= - sin^{3} (θ) + 3 cos^{2} (θ) sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (x) sin (3 x) = 4 sin (x) cos^{2} (x)

p ro v e sec (\frac{π}{2} - t) = csc (t)

p ro v e sin (3 u) = sin (u) (3 - 4 sin^{2} (u))

p ro v e \frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )} = sin (A) + cos (A)

p ro v e cos (x - \frac{π}{6}) - sin (x + \frac{π}{3}) = 0

Is sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ) ?
The answer to whether sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ) is True